Question

if ( f(x)=\frac{3 x^{2}+4 x + 7}{sqrt{x}} ), then:
( f^{prime}(x)= )
( f^{prime}(2)= )

Explanation:

Step1: Simplify the function

Rewrite \(f(x)=\frac{3x^{2}+4x + 7}{\sqrt{x}}\) as \(f(x)=3x^{\frac{3}{2}}+4x^{\frac{1}{2}}+7x^{-\frac{1}{2}}\) using the rule \(\frac{a^{m}}{a^{n}}=a^{m - n}\) (\(a=x\), \(m = 2,1,0\) and \(n=\frac{1}{2}\)).

Step2: Differentiate term - by - term

Use the power rule \((x^{n})^\prime=nx^{n - 1}\).
For \(y = 3x^{\frac{3}{2}}\), \(y^\prime=3\times\frac{3}{2}x^{\frac{3}{2}-1}=\frac{9}{2}x^{\frac{1}{2}}\).
For \(y = 4x^{\frac{1}{2}}\), \(y^\prime=4\times\frac{1}{2}x^{\frac{1}{2}-1}=2x^{-\frac{1}{2}}\).
For \(y = 7x^{-\frac{1}{2}}\), \(y^\prime=7\times(-\frac{1}{2})x^{-\frac{1}{2}-1}=-\frac{7}{2}x^{-\frac{3}{2}}\).
So \(f^\prime(x)=\frac{9}{2}\sqrt{x}+\frac{2}{\sqrt{x}}-\frac{7}{2x\sqrt{x}}\).

Step3: Evaluate \(f^\prime(2)\)

Substitute \(x = 2\) into \(f^\prime(x)\).
\(f^\prime(2)=\frac{9}{2}\sqrt{2}+\frac{2}{\sqrt{2}}-\frac{7}{2\times2\times\sqrt{2}}\).
Rationalize the denominators: \(\frac{2}{\sqrt{2}}=\sqrt{2}\), \(\frac{7}{4\sqrt{2}}=\frac{7\sqrt{2}}{8}\).
\(f^\prime(2)=\frac{9\sqrt{2}}{2}+\sqrt{2}-\frac{7\sqrt{2}}{8}\).
Find a common denominator (\(8\)): \(\frac{9\sqrt{2}\times4}{2\times4}+\frac{\sqrt{2}\times8}{1\times8}-\frac{7\sqrt{2}}{8}=\frac{36\sqrt{2}+16\sqrt{2}-7\sqrt{2}}{8}=\frac{45\sqrt{2}}{8}\).

Answer:

\(f^\prime(x)=\frac{9}{2}\sqrt{x}+\frac{2}{\sqrt{x}}-\frac{7}{2x\sqrt{x}}\); \(f^\prime(2)=\frac{45\sqrt{2}}{8}\)